我正在构建遗传算法以在python中进行特征选择。我从我的数据中提取了特征，然后我分成了两个数据帧，“训练”和“测试”数据帧。 如何在“人口”数据框（每个单独）和“训练”数据框中为每一行复用多个值？
'train'数据框：
feature0 feature1 feature2 feature3 feature4 feature5
0 18.279579 -3.921346 13.611829 -7.250185 -11.773605 -18.265003
1 17.899545 -15.503942 -0.741729 -0.053619 -6.734652 4.398419
4 16.432750 -22.490190 -4.611659 -15.247781 -13.941488 -2.433374
5 15.905368 -4.812785 18.291712 3.742221 3.631887 -1.074326
6 16.991823 -15.946251 8.299577 8.057511 8.057510 -1.482333
'人口'数据框：
0 1 2 3 4 5
0 1 1 0 0 0 1
1 0 1 0 1 0 0
2 0 0 0 0 0 1
3 0 0 1 0 1 1
将'population'中的每一行乘以'train'中的所有行。 结果如下：
1）来自人口第1行：
feature0 feature1 feature2 feature3 feature4 feature5
0 18.279579 -3.921346 0 0 0 -18.265003
1 17.899545 -15.503942 0 0 0 4.398419
4 16.432750 -22.490190 0 0 0 -2.433374
5 15.905368 -4.812785 0 0 0 -1.074326
6 16.991823 -15.946251 0 0 0 -1.482333
2）来自人口第2行：
feature0 feature1 feature2 feature3 feature4 feature5
0 0 -3.921346 0 -7.250185 0 0
1 0 -15.503942 0 -0.053619 0 0
4 0 -22.490190 0 -15.247781 0 0
5 0 -4.812785 0 3.742221 0 0
6 0 -15.946251 0 8.057511 0 0
等等......
答案 0 :(得分：4)
如果需要循环（如果大数据则缓慢）：
for i, x in population.iterrows():
print (train * x.values)
feature0 feature1 feature2 feature3 feature4 feature5
0 18.279579 -3.921346 0.0 -0.0 -0.0 -18.265003
1 17.899545 -15.503942 -0.0 -0.0 -0.0 4.398419
4 16.432750 -22.490190 -0.0 -0.0 -0.0 -2.433374
5 15.905368 -4.812785 0.0 0.0 0.0 -1.074326
6 16.991823 -15.946251 0.0 0.0 0.0 -1.482333
feature0 feature1 feature2 feature3 feature4 feature5
0 0.0 -3.921346 0.0 -7.250185 -0.0 -0.0
1 0.0 -15.503942 -0.0 -0.053619 -0.0 0.0
4 0.0 -22.490190 -0.0 -15.247781 -0.0 -0.0
5 0.0 -4.812785 0.0 3.742221 0.0 -0.0
6 0.0 -15.946251 0.0 8.057511 0.0 -0.0
feature0 feature1 feature2 feature3 feature4 feature5
0 0.0 -0.0 0.0 -0.0 -0.0 -18.265003
1 0.0 -0.0 -0.0 -0.0 -0.0 4.398419
4 0.0 -0.0 -0.0 -0.0 -0.0 -2.433374
5 0.0 -0.0 0.0 0.0 0.0 -1.074326
6 0.0 -0.0 0.0 0.0 0.0 -1.482333
feature0 feature1 feature2 feature3 feature4 feature5
0 0.0 -0.0 13.611829 -0.0 -11.773605 -18.265003
1 0.0 -0.0 -0.741729 -0.0 -6.734652 4.398419
4 0.0 -0.0 -4.611659 -0.0 -13.941488 -2.433374
5 0.0 -0.0 18.291712 0.0 3.631887 -1.074326
6 0.0 -0.0 8.299577 0.0 8.057510 -1.482333
或者每一行分开：
print (train * population.values[0])
feature0 feature1 feature2 feature3 feature4 feature5
0 18.279579 -3.921346 0.0 -0.0 -0.0 -18.265003
1 17.899545 -15.503942 -0.0 -0.0 -0.0 4.398419
4 16.432750 -22.490190 -0.0 -0.0 -0.0 -2.433374
5 15.905368 -4.812785 0.0 0.0 0.0 -1.074326
6 16.991823 -15.946251 0.0 0.0 0.0 -1.482333
或者对于MultiIndex DataFrame：
d = pd.concat([train * population.values[i] for i in range(population.shape[0])],
keys=population.index.tolist())
print (d)
feature0 feature1 feature2 feature3 feature4 feature5
0 0 18.279579 -3.921346 0.000000 -0.000000 -0.000000 -18.265003
1 17.899545 -15.503942 -0.000000 -0.000000 -0.000000 4.398419
4 16.432750 -22.490190 -0.000000 -0.000000 -0.000000 -2.433374
5 15.905368 -4.812785 0.000000 0.000000 0.000000 -1.074326
6 16.991823 -15.946251 0.000000 0.000000 0.000000 -1.482333
1 0 0.000000 -3.921346 0.000000 -7.250185 -0.000000 -0.000000
1 0.000000 -15.503942 -0.000000 -0.053619 -0.000000 0.000000
4 0.000000 -22.490190 -0.000000 -15.247781 -0.000000 -0.000000
5 0.000000 -4.812785 0.000000 3.742221 0.000000 -0.000000
6 0.000000 -15.946251 0.000000 8.057511 0.000000 -0.000000
2 0 0.000000 -0.000000 0.000000 -0.000000 -0.000000 -18.265003
1 0.000000 -0.000000 -0.000000 -0.000000 -0.000000 4.398419
4 0.000000 -0.000000 -0.000000 -0.000000 -0.000000 -2.433374
5 0.000000 -0.000000 0.000000 0.000000 0.000000 -1.074326
6 0.000000 -0.000000 0.000000 0.000000 0.000000 -1.482333
3 0 0.000000 -0.000000 13.611829 -0.000000 -11.773605 -18.265003
1 0.000000 -0.000000 -0.741729 -0.000000 -6.734652 4.398419
4 0.000000 -0.000000 -4.611659 -0.000000 -13.941488 -2.433374
5 0.000000 -0.000000 18.291712 0.000000 3.631887 -1.074326
6 0.000000 -0.000000 8.299577 0.000000 8.057510 -1.482333
并按xs选择：
print (d.xs(0))
feature0 feature1 feature2 feature3 feature4 feature5
0 18.279579 -3.921346 0.0 -0.0 -0.0 -18.265003
1 17.899545 -15.503942 -0.0 -0.0 -0.0 4.398419
4 16.432750 -22.490190 -0.0 -0.0 -0.0 -2.433374
5 15.905368 -4.812785 0.0 0.0 0.0 -1.074326
6 16.991823 -15.946251 0.0 0.0 0.0 -1.482333
答案 1 :(得分：2)
我使用numpy广播一次性完成所有操作......
train_ = pd.DataFrame(
(train.values * pop.values[:, None]).reshape(-1, train.shape[1]),
pd.MultiIndex.from_product([pop.index, train.index]),
train.columns
)
train_
feature0 feature1 feature2 feature3 feature4 feature5
0 0 18.279579 -3.921346 0.000000 -0.000000 -0.000000 -18.265003
1 17.899545 -15.503942 -0.000000 -0.000000 -0.000000 4.398419
4 16.432750 -22.490190 -0.000000 -0.000000 -0.000000 -2.433374
5 15.905368 -4.812785 0.000000 0.000000 0.000000 -1.074326
6 16.991823 -15.946251 0.000000 0.000000 0.000000 -1.482333
1 0 0.000000 -3.921346 0.000000 -7.250185 -0.000000 -0.000000
1 0.000000 -15.503942 -0.000000 -0.053619 -0.000000 0.000000
4 0.000000 -22.490190 -0.000000 -15.247781 -0.000000 -0.000000
5 0.000000 -4.812785 0.000000 3.742221 0.000000 -0.000000
6 0.000000 -15.946251 0.000000 8.057511 0.000000 -0.000000
2 0 0.000000 -0.000000 0.000000 -0.000000 -0.000000 -18.265003
1 0.000000 -0.000000 -0.000000 -0.000000 -0.000000 4.398419
4 0.000000 -0.000000 -0.000000 -0.000000 -0.000000 -2.433374
5 0.000000 -0.000000 0.000000 0.000000 0.000000 -1.074326
6 0.000000 -0.000000 0.000000 0.000000 0.000000 -1.482333
3 0 0.000000 -0.000000 13.611829 -0.000000 -11.773605 -18.265003
1 0.000000 -0.000000 -0.741729 -0.000000 -6.734652 4.398419
4 0.000000 -0.000000 -4.611659 -0.000000 -13.941488 -2.433374
5 0.000000 -0.000000 18.291712 0.000000 3.631887 -1.074326
6 0.000000 -0.000000 8.299577 0.000000 8.057510 -1.482333
您只能访问与第i行对应的那个或population与train_.loc[i]
train_.loc[3]
feature0 feature1 feature2 feature3 feature4 feature5
0 0.0 -0.0 13.611829 -0.0 -11.773605 -18.265003
1 0.0 -0.0 -0.741729 -0.0 -6.734652 4.398419
4 0.0 -0.0 -4.611659 -0.0 -13.941488 -2.433374
5 0.0 -0.0 18.291712 0.0 3.631887 -1.074326
6 0.0 -0.0 8.299577 0.0 8.057510 -1.482333
粗略的时间测试
我懒得做更强大的测试
%%timeit
pd.DataFrame(
(train.values * pop.values[:, None]).reshape(-1, train.shape[1]),
pd.MultiIndex.from_product([pop.index, train.index]),
train.columns
)
%%timeit
res = pop.iloc[np.repeat(np.arange(len(pop)), len(train))]
res = res.set_index(np.tile(train.index, len(pop)), append=True).add_prefix('feature')
res.mul(train, level=1)
%%timeit
pd.concat([train * pop.values[i] for i in range(pop.shape[0])],
keys=pop.index.tolist())
571 µs ± 10.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
1.42 ms ± 18 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
1.7 ms ± 69.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
答案 2 :(得分：1)
将population的列设置为与train匹配后，您可以使用*：
In [11]: population.columns = train.columns
In [12]: train * population.iloc[0]
Out[12]:
feature0 feature1 feature2 feature3 feature4 feature5
0 18.279579 -3.921346 0.0 -0.0 -0.0 -18.265003
1 17.899545 -15.503942 -0.0 -0.0 -0.0 4.398419
4 16.432750 -22.490190 -0.0 -0.0 -0.0 -2.433374
5 15.905368 -4.812785 0.0 0.0 0.0 -1.074326
6 16.991823 -15.946251 0.0 0.0 0.0 -1.482333
您可以使用np.tile和np.repeat非常有效地制作MultiIndex（由@jezrael推荐）：
In [11]: res = population.iloc[np.repeat(np.arange(len(population)), len(train))]
In [12]: res = res.set_index(np.tile(train.index, len(population)), append=True)
In [13]: res
Out[13]:
feature0 feature1 feature2 feature3 feature4 feature5
0 0 1 1 0 0 0 1
1 1 1 0 0 0 1
4 1 1 0 0 0 1
5 1 1 0 0 0 1
6 1 1 0 0 0 1
1 0 0 1 0 1 0 0
1 0 1 0 1 0 0
4 0 1 0 1 0 0
5 0 1 0 1 0 0
6 0 1 0 1 0 0
2 0 0 0 0 0 0 1
1 0 0 0 0 0 1
4 0 0 0 0 0 1
5 0 0 0 0 0 1
6 0 0 0 0 0 1
3 0 0 0 1 0 1 1
1 0 0 1 0 1 1
4 0 0 1 0 1 1
5 0 0 1 0 1 1
6 0 0 1 0 1 1
In [14]: res.mul(train, level=1)
Out[14]:
feature0 feature1 feature2 feature3 feature4 feature5
0 0 18.279579 -3.921346 0.000000 -0.000000 -0.000000 -18.265003
1 17.899545 -15.503942 -0.000000 -0.000000 -0.000000 4.398419
4 16.432750 -22.490190 -0.000000 -0.000000 -0.000000 -2.433374
5 15.905368 -4.812785 0.000000 0.000000 0.000000 -1.074326
6 16.991823 -15.946251 0.000000 0.000000 0.000000 -1.482333
1 0 0.000000 -3.921346 0.000000 -7.250185 -0.000000 -0.000000
1 0.000000 -15.503942 -0.000000 -0.053619 -0.000000 0.000000
4 0.000000 -22.490190 -0.000000 -15.247781 -0.000000 -0.000000
5 0.000000 -4.812785 0.000000 3.742221 0.000000 -0.000000
6 0.000000 -15.946251 0.000000 8.057511 0.000000 -0.000000
2 0 0.000000 -0.000000 0.000000 -0.000000 -0.000000 -18.265003
1 0.000000 -0.000000 -0.000000 -0.000000 -0.000000 4.398419
4 0.000000 -0.000000 -0.000000 -0.000000 -0.000000 -2.433374
5 0.000000 -0.000000 0.000000 0.000000 0.000000 -1.074326
6 0.000000 -0.000000 0.000000 0.000000 0.000000 -1.482333
3 0 0.000000 -0.000000 13.611829 -0.000000 -11.773605 -18.265003
1 0.000000 -0.000000 -0.741729 -0.000000 -6.734652 4.398419
4 0.000000 -0.000000 -4.611659 -0.000000 -13.941488 -2.433374
5 0.000000 -0.000000 18.291712 0.000000 3.631887 -1.074326
6 0.000000 -0.000000 8.299577 0.000000 8.057510 -1.482333